Question

A 144-Ω light bulb is connected to a conducting wire that is wrapped into the shape of a square with side length of 83.0 cm. This square loop is rotated within a uniform magnetic field of 454 mT. What is the change in magnetic flux through the loop when it rotates from a position where its area vector makes an angle of 30° with the field to a position where the area vector is parallel to the field?

A 144-Ω light bulb is connected to a conducting wire that is wrapped into the shape of a square with side length of 83.0 cm. This square loop is rotated within a uniform magnetic field of 454 mT. The loop rotates from a position where its area vector makes an angle of 30° with the field to a position where the area vector is parallel to the field in 56.3 ms. What is the induced current through the light bulb?

A 144-Ω light bulb is connected to a conducting wire that is wrapped into the shape of a square with side length of 83.0 cm. This square loop is rotated with a frequency of 60 Hz within a uniform magnetic field of 454 mT. This means the loop makes half a revolution in 8.33 ms. What is the induced current in the light bulb when the loop rotates from a position where its area vector is opposite the magnetic field to a position where its area vector is parallel to the magnetic field?

Answer 1

1] When the area vector is parallel with the magnetic field, the magnetic flux is:

$$ \phi_{1}=B A=B L^{2} $$

when the area vector makes an angle of \(30^{\circ}\) with respect to the magnetic field axis, the magnetic flux is:

$$ \phi_{2}=B A \cos 30^{\circ}=B L^{2} \cos 30^{\circ} $$

so, change in magnetic flux will be:

$$ \phi_{2}-\phi_{1}=B L^{2}\left(1-\cos 30^{\circ}\right)=\left(454 \times 10^{-3}\right)(0.83)^{2}\left(1-\cos 30^{\circ}\right)=0.0419 W b $$

2] From above, the change in magnetic flux is:

$$ \Delta \phi=0.0419 W b $$

this change occurs in \(56.3 \mathrm{~ms}\)

and from Faraday's law of electromagnetic induction, the emf induced is the rate of change of magnetic flux with time.

So, induced current \(=\mathrm{emf} /\) resistance

$$ \Rightarrow \quad i=\frac{V}{R}=\frac{1}{R} \frac{\Delta \phi}{\Delta t}=\frac{1}{(144)} \frac{(0.0419)}{56.3 \times 10^{-3}}=5.168 \mathrm{~mA} $$

3] The magnitude of induced current will now be:

$$ i=\frac{V}{R}=\frac{1}{R} \frac{d \phi}{d t}=\frac{1}{R} B A \omega $$

$$ \Rightarrow i=\frac{1}{144}\left(454 \times 10^{-3}\right)\left(0.83^{2}\right)(2 \pi[60])=0.8188 A \text {. } $$